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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper II covers organic chemistry and spectroscopy.

UPSC Civil Services (Mains), Chemistry optional · 2018 · Paper II

Q3(a). Two transformations: indole and 1-vinylcyclohexene

  • 10 marks
  • Answered
Question

Carry out the following transformations:
(i) [structure] → [structure]
(ii) [structure] → [structure]

Answer in brief

(i) N-sulfonylation, C-2 lithiation, ethylene oxide, then N-deprotection. (ii) Cyclic hydroboration with thexylborane, carbonylation with CO and oxidation with H₂O₂/OH⁻: trans-octahydro-1H-inden-1-one.

Key points

  • N-phenylsulfonylindole. The N–H is removed so that the base does not simply deprotonate N, and the sulfonyl group directs lithiation to C-2 by coordination and induction.
  • 2-lithio-1-(phenylsulfonyl)indole (C-2 H is the most acidic ring C–H).
  • SN2 ring opening at the epoxide CH₂ gives the 2-(2-hydroxyethyl) side chain.
  • removes the sulfonyl group (N–S cleavage), giving the product.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Reading of the question: the Hindi text asks for the mechanism, the English for the transformation; the reagents of each step are given with the key mechanism.

Introduction. (i) Electrophiles attack indole at C-3, so a C-2 substituent needs a different approach: protect N, lithiate C-2 (directed metalation), and use ethylene oxide as a two-carbon electrophile. (ii) The product has one more carbon than the diene (C₈H₁₂ → C₉H₁₄O); a cyclic hydroboration of both C=C bonds followed by carbonylation (Brown) stitches in the C=O as the ring carbon.

(i) Indole → 2-(1H-indol-2-yl)ethan-1-ol

  1. NaH, then PhSO2Cl (THF): N-phenylsulfonylindole. The N–H is removed so that the base does not simply deprotonate N, and the sulfonyl group directs lithiation to C-2 by coordination and induction.
  2. LDA (or n-BuLi), THF, -78°C: 2-lithio-1-(phenylsulfonyl)indole (C-2 H is the most acidic ring C–H).
  3. Ethylene oxide (with BF₃.OEt₂), then H₃O⁺: SN2 ring opening at the epoxide CH₂ gives the 2-(2-hydroxyethyl) side chain.
  4. NaOH or KOH in methanol, reflux: removes the sulfonyl group (N–S cleavage), giving the product.

Scheme: Directed C-2 lithiation, epoxide opening, N-deprotection.

(ii) 1-Vinylcyclohexene → octahydro-1H-inden-1-one (hydrindan-1-one)

  1. Thexylborane (ThxBH₂, from BH₃ and 2,3-dimethylbut-2-ene), THF, 0°C. The first B–H adds to the terminal vinyl CH₂; the second adds intramolecularly to the ring C=C, B to C-2 (anti-Markovnikov), syn. A five-membered boracycle (borolane) fused to the cyclohexane is formed.
  2. Carbon monoxide (about 70 atm, 50°C, in the presence of water): CO adds to boron, and both ring carbons migrate in turn from B to the CO carbon (the thexyl group migrates last and is kept back).
  3. Alkaline H₂O₂: oxidises the C–B bonds, releasing the ketone. The CO carbon is now the C=O between the old vinyl CH₂ and ring C-2: a cyclopentanone fused to the cyclohexane.

Scheme: Each B–C bond migrates with retention, so the trans ring fusion of the borane is kept.

Stereochemistry: hydroboration places H (on C-3a) and B (on C-7a) cis; the two ring-fusion H atoms are therefore trans, and migration with retention gives trans-hydrindan-1-one.

Conclusion. (i) N-sulfonylation, C-2 lithiation, ethylene oxide, then N-deprotection. (ii) Cyclic hydroboration with thexylborane, carbonylation with CO and oxidation with H₂O₂/OH⁻: trans-octahydro-1H-inden-1-one.

Application. Directed lithiation is how C-2 substituted indoles (tryptamine and indole alkaloid analogues) are made; Brown's diene-to-cyclic-ketone route builds the hydrindanone (C/D ring) unit of steroids.

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Model answer to Q3(a), 2018 Paper II: book edition page 11
Book edition, page 11
Model answer to Q3(a), 2018 Paper II: book edition page 12
Book edition, page 12
Model answer to Q3(a), 2018 Paper II: book edition page 13
Book edition, page 13